Gravitational Acceleration at Different Latitudes

[duggielanger]

Homework Statement

The rotation of the Earth affects the apparent gravitational acceleration at
different latitudes. At a location on the Equator, a man’s weight is registered
as 709.7 N on a set of very accurate scales. Assuming a perfectly spherical
Earth, determine what the same set of scales would register for the same
man’s weight at the North Pole

Homework Equations

Not sure I have used f=ma and w=mg but don't think I need these as the Earth is perfectly spherical
So it might be a=v^2/r

The Attempt at a Solution

As far as I can figure if Earth is perfectly spherical then there will be no change in weight due to the force of gravity. But maybe because of the rotation and Centrifugal force this might have a slight effect.

 

[SammyS]

Homework Statement

The rotation of the Earth affects the apparent gravitational acceleration at
different latitudes. At a location on the Equator, a man’s weight is registered
as 709.7 N on a set of very accurate scales. Assuming a perfectly spherical
Earth, determine what the same set of scales would register for the same
man’s weight at the North Pole

Homework Equations

Not sure I have used f=ma and w=mg but don't think I need these as the Earth is perfectly spherical
So it might be a=v^2/r

The Attempt at a Solution

As far as I can figure if Earth is perfectly spherical then there will be no change in weight due to the force of gravity. But maybe because of the rotation and Centrifugal force this might have a slight effect.

Correct.
So, how much effect ?

 

[duggielanger]

So if I find the angular velocity first ω=2∏rad/T and then v=rω^2
Plugged in values
2*3.14/(24h*3600s/h)=7.3x10^-5 rads
v=6.378x10^6m/(7.3x10^-5)^2rads=0.034m/s^2
So the force of gravity will be 0.034m/s^2 less
and then I can find the weight .
Dose this seem right

 

[SammyS]

So if I find the angular velocity first ω=2∏rad/T and then v=rω^2
Plugged in values
2*3.14/(24h*3600s/h)=7.3x10^-5 rads/s²
v=6.378x10^6m/(7.3x10^-5)^2rads/s²=0.034m/s^2
So the [STRIKE]force[/STRIKE] acceleration of gravity will be 0.034m/s^2 less
and then I can find the weight .
Does this seem right

Yes, except for some units and saying force rather than acceleration .

(Corrected above)

 

[duggielanger]

Ok thanks A lot SammyS your help is appreciated

 

[SammyS]

And ... I forgot to say,

Welcome to PF !